Exam 3

Question 1

An n × n matrix A is diagonalizable if there exist n × n matrices D and P, with D diagonal and P invertible, such that
A = PDP−1

Answer 1

Diagonalizable

Question 2

A vector space consists of a nonempty set V of vectors together with operations of addition and scalar multiplication on the vectors that satisfy each of the following:

(1) If v1 and v2 are in V, then so is v1 + v2. Hence V is closed under addition.
(2) If c is a real scalar and v is in V, then so is cv. Hence V is closed under scalar multiplication.
(3) There exists a zero vector 0 in V such that 0+v = v for all v in V.
(4) For each v in V there exists an additive inverse (or opposite) vector −v in V such that v + (−v) = 0 for all v in V.
(5) For all v1, v2, and v3 in V and real scalars c1 and c2, we have
(a) v1 + v2 = v2 + v1
(b) (v1 + v2) + v3 = v1 + (v2 + v3)
(c) c1(v1 + v2) = c1v1 + c1v2
(d) (c1 + c2)v1 = c1v1 + c2v1
(e) (c1c2)v1 = c1(c2v1)
(f) 1 · v1 = v1

Answer 2

Vector Space

Question 3

A subset S of vectors is a subspace if S satisfies the following three conditions:

(1) S contains 0, the zero vector,
(2) if u and v are in S, then u + v is also in S, and
(3) if r is a real number and u is in S, then ru is also in S

Answer 3

Subspace

Question 4

Let = {v1, v2, . . . , vm} be a nonempty set of vectors in a vector space V. The span of this set is denoted span{v1, v2, . . . , vm} and is defined to be the set of all linear combinations of form
c1v1 + c2v2 + + cmvm

where c1, c2, . . . , cm can be any real numbers.
If consists of infinitely many vectors, then we define span() to be the set of all linear combinations of finite subsets of .

Answer 4

Span of a vector space

Question 5

Let V = {v1, v2, . . . , vm} be a set of vectors in a vector space V. Then V is linearly independent if the equation ,
c1 v1 + c2v2 + + cmvm = 0
has only the trivial solution c1 = . . . = cm = 0. The set is linearly dependent if the equation has any nontrivial solutions.

Answer 5

Linearly independent/dependent for a vector space

Question 6

Let (V) be a subset of a vector space V. Then (V) is a basis of V if (V) is linearly independent and spans V

Answer 6

Basis

Question 7

Let V and W be vector spaces. Then T : V → W is a linear transformation if for all v1 and v2 in V and all real scalars c, the function T satisfies

(a) T(v1 + v2) = T(v1) + T(v2)
(b) T(c v1) = cT(v1)

Answer 7

Linear Transformation for vector space

Question 8

If v is a vector in V, then T(v) is the image of v under T

Answer 8

Image

Question 9

If S is a subspace of V, then T(S) denotes the subset of W consisting of all images of elements of S

Answer 9

Elements

Question 10

The Range of T is denoted range(T) and is the subset of W consisting of all images of elements of V

Answer 10

Range of T

Question 11

The kernel of T is denoted Ker(T) and is the set of all elements v in V such that T(v)=0

Answer 11

Kernel of T

Question 12

Let T : V → W be a linear transformation. Then

(a) T is one-to-one if for each w in W there is atmost one v in V such that T(v) = w.
(b) T is onto if for each w in W there is at least one v in V such that T(v) = w.

Answer 12

One-to-one and Onto

Question 13

A linear transformation T : V → W is an isomorphism if T is both one-to-one and onto. If such an isomorphism exists, then we say that V and W are isomorphic vector spaces.

Answer 13

Isomorphism

Question 14

A linear transformation T : V → W is invertible if T is one-to-one and onto. When T is invertible, the inverse function T−1 : W → V is defined by
T−1(w) = v if and only if T(v) = w

Answer 14

Inverse/Invertible